Question

In a container 96g O2 and 335g Ne is present. If total pressure of gaseous mixture is 25 atm. Then what will be partial pressure of Ne?

Answer 1

Given :- 96g of O2 and 335g of Ne

Partial pressure of Ne = Mole fraction of Ne * Total pressure

Pp of Ne = (16.75/19.75)*25

= 21 approx

Answer 2

Answer:

The partial pressure of Ne gas is 21 atm.

Explanation:

Mass of oxygen gas= 96 g

moles of oxygen gas= $n_o=\frac{96 g}{32 g/mol}=3 mol$

Mass of neon gas =335 g

Moles of neon gas =$n_n=\frac{335 g}{20 g/mol}=16.75 mol$

Mole fraction of neon gas =$\frac{n_n}{n_o+n_n}=\frac{16.75 mol}{3 mol+16.75 mol}=0.84$

Total pressure of gaseous mixture = 25 atm

Partial pressure of neon gas :

$p_n=P\times \chi_{Ne}=25 atm\times 0.84=21 atm$

The partial pressure of Ne gas is 21 atm.